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08 Dec 2020, 03:45
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C
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Difficulty:
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Question Stats:
55% (02:46) correct
45%
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wrong
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In a green view apartment, the houses of a row are numbered consecutively from 1 to 49. Assuming that there is a value of 'x' such that the sum of the numbers of the houses preceding the house numbered 'x' is equal to the sum of the numbers of the houses following it. Then, what will be the value of 'x'?
A. 21
B. 34
C. 35
D. 36
E. 43
A. 21
B. 34
C. 35
D. 36
E. 43
08 Dec 2020, 05:04
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CONCEPT:
Sum of the first n natural numbers =n(n+1)/2
SOLUTION:
-According to the question-
Sum of the first (x-1) natural numbers = Sum of natural number from (x+1) to 49.
So,
(x−1)(x)/2=(49×50)/2 - (x)(x+1)/2
⇒ 2x^2=49×50
⇒ x=7×5= 35 (C)
Hope this helps
Devmitra Sen
Sum of the first n natural numbers =n(n+1)/2
SOLUTION:
-According to the question-
Sum of the first (x-1) natural numbers = Sum of natural number from (x+1) to 49.
So,
(x−1)(x)/2=(49×50)/2 - (x)(x+1)/2
⇒ 2x^2=49×50
⇒ x=7×5= 35 (C)
Hope this helps
Devmitra Sen
16 Jul 2022, 16:50
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Bunuel
Let's just try C. If it's right, great. If it's way too high, we know it's A. If it's a tiny bit too high, we know it's B. If it's a tiny bit too low, we know it's D. If it's way too low, we know it's E.
1+2+3+4...+34 = 35*17 = 595
36+37+38...+49 = 85*7 = 595
Yay!
Answer choice C.
It's a pretty straightforward question, so about another way, too?
The units digits of each ten numbers sum to 45. There are five set of ten numbers, so the units digit of the sum will be 5. In order to have the two halves be equal, we need to remove an odd number from our list. B and D are wrong. A is obviously wrong; how would 1 through 20 sum to the same as more numbers each of which is larger than the largest in 1 through 20? 43 is obviously wrong; that only leave 44+45+46+47+48+49 and just the 30's are going to be more than that. Only C is left.
